Lehrstuhl für Mechanische Verfahrenstechnik 1 Prof. Dr.-Ing. Jürgen Tomas

Lecture: Mechanical Process Engineering

Exercise – Flow separation in a hydrocyclone

Task:

A suspension of kaolin (china clay/ porcelain earth) (20°C) with a solid concentration 𝑐𝑆= 250

kg/m³ and a solid mass flow rate of the feed �̇�𝑆 = 100 t/h shall be classified (𝑑𝑇 = 30 μm).

The product contains pure porcelain earth 𝜌𝐾 = 2580 kg/m³ and quartz 𝜌𝑄 = 2650 kg/m³. The

average solid density is 𝜌𝑆 = 2600 kg/m³. In the fine product there shall be a higher concen-

tration of pure kaolin and in the coarse product a higher concentration of quartz.

It is known, that the product suspension (feed) has a GGS-particle size distribution, with the

parameters 𝑑80 = 31.3 μm and 𝑘 = 0.7.

The suspension runs from a barrel to a hydrocyclone by use of gravity. The barrel is put on a

level (ℎ) 15 m higher than the cyclone feed opening.

Four different cyclones with diameters 𝐷 of 50, 100, 200, 400 mm, and a cone angel 𝛼 = 20°

are available.

Given:

solid concentration 𝑐𝑆 = 250 kg/m³

cut particle size 𝑑𝑇 = 30 µm

mass flow rate �̇�𝑆 = 100 t/h

average solid density 𝜌𝑆 = 2600 kg/m³

density kaolinit 𝜌𝐾 = 2580 kg/m³

density quarz 𝜌𝑄 = 2650 kg/m³

density water (20°C) 𝜌𝐹 = 998 kg/m³

dynamic viscosity of water 𝜂𝐷,𝐹 = 1 mPas

Wanted:

Number of hydrocyclones 𝑛𝑍𝑦𝑘𝑙

𝐷

𝐷𝑜

𝐷𝑖

�̇�𝑜 = �̇�𝐹 + �̇�𝑓𝑙𝑢𝑖𝑑,𝑜

�̇�𝑆𝑢𝑠𝑝

𝐷𝑎

�̇�𝑎 = �̇�𝐺 + �̇�𝑓𝑙𝑢𝑖𝑑,𝑎

Lehrstuhl für Mechanische Verfahrenstechnik 2 Prof. Dr.-Ing. Jürgen Tomas

Basics:

The Reynolds numbers 𝑅𝑒 in hydrocyclone-flows are between 𝑅𝑒 = 105 to 106, so there are

high turbulent flow conditions.

modified model of the turbulent cross-flow classification is requiered

Calculation of the inlet nozzles 𝐷𝑖, the overflow nozzle diameter 𝐷𝑜 and the upper outlet noz-

zles 𝐷𝑎.

𝐷𝑖 = 0.2 ∙ 𝐷 (1)

𝐷𝑜 = 0.3 ∙ 𝐷 (2)

𝐷𝑎 =2

3∙ 𝐷𝑜 (3)

simple empirical relations for low concentrated slurries (𝜑𝑝= 5 bis 10 %)

a) according to PLITT

�̇�𝑜

�̇�𝑎= (

𝐷𝑜

𝐷𝑎)3 bis 4

(4)

b) according to TARJÁN

�̇�𝑜

�̇�𝑎≈ 0.91 ∙ (

𝐷𝑜

𝐷𝑎)3

(5)

Calculation of the cut size diameter 𝑑𝑇 (includes TARJÁN)

𝑑𝑇 = 0.284 ∙ 𝑘𝑑 ∙ √

𝜂 ∙ 𝐷 ∙ ln [0.91 ∙ (𝐷𝑜𝐷𝑎

)3

]

(1 − 𝜑𝑆)𝑛 ∙ (𝜌𝑆 − 𝜌𝐹) ∙ √

𝑝𝑖𝜌𝑆𝑢𝑠𝑝

(6)

with:

𝑘𝑑 = [220 ∙ 𝑑50 ∙ √𝜌𝑆 − 𝜌𝐹

𝐷]

𝑚

(7)

with 𝑚 = {5 ∙ 𝐷 für 𝐷 < 0.1 𝑚0,5 für 𝐷 ≥ 0.1 𝑚

The transformation of equation (6) gives for the diameter ratio:

𝐷𝑜

𝐷𝑎= √

1

0.91∙ exp

[

(𝑑𝑇

0.284 ∙ 𝑘𝑑)2

∙

(1 − 𝜑𝑆)3 ∙ (𝜌𝑆 − 𝜌𝐹) ∙ √

𝑝𝑖𝜌𝑆𝑢𝑠𝑝

𝜂 ∙ 𝐷

] 3

(8)

Lehrstuhl für Mechanische Verfahrenstechnik 3 Prof. Dr.-Ing. Jürgen Tomas

Calculation of a theoretical separation sharpness of a hydrocyclone

𝜅 =𝑑25

𝑑75=

[ ln [0.303 ∙ (

𝐷𝑜𝐷𝑎

)3

]

ln [2.73 ∙ (𝐷𝑜𝐷𝑎

)3

]]

12

(9)

The suspension flow rate of a hydrocyclone �̇�𝑍𝑦𝑘𝑙 can be calculated by the following formula.

�̇�𝑍𝑦𝑘𝑙 = 𝑘𝛼 ∙ 𝐷𝑖 ∙ 𝐷𝑜 ∙ √𝑝𝑖

𝜌𝑆𝑢𝑠𝑝 (10)

with

𝑘𝛼 =1

3.6 für 𝛼 = 20° (11)

or:

𝑘𝛼 =0.225

𝛼𝐵𝑜𝑔𝑒𝑛0.2 𝛼 in radian (𝛼𝐵𝑜𝑔𝑒𝑛 = 𝛼𝐺𝑟𝑎𝑑 ∙

𝜋

180) (12)

Solution:

Calculation of each diameter 𝐷𝑖, 𝐷𝑜, 𝐷𝑎 and characteristics 𝑘𝑑 and 𝑚 (see Table 1) according

to equation (1) to (3) and (7).

For the determination of 𝑘𝑑 the 𝑑50 diameter of the GGS-distribution is used by BOND.

𝑄3(𝑑) = 0.8 ∙ (𝑑

𝑑80)𝑘

(13)

After conversion of equation (13) and with 𝑄3(𝑑) = 0.5; 𝑘 = 0.7 and 𝑑80 = 31.3 µm results for

the 𝑑50 diameter:

𝑑50 = 31.3 µm ∙ √0.5

0.8

0.7

= 16 µm

Table 1: solution table I

hydrocyclone

𝐷 in m

inlet nozzle

𝐷𝑖 in m

overflow nozzle

𝐷𝑜 in m

underflow nozzle

𝐷𝑎 in m 𝑘𝑑 𝑚

0.05 0.01 0.015 0.01 0.891 0.25

0.1 0.02 0.03 0.02 0.667 0.5

0.2 0.04 0.06 0.04 0.561 0.5

0.4 0.08 0.12 0.08 0.471 0.5

Lehrstuhl für Mechanische Verfahrenstechnik 4 Prof. Dr.-Ing. Jürgen Tomas

For the calculation of the diameter ratio according to equation (8) it's important to determine

the volume fraction of solids in the feed stream 𝜑𝑆 by the solid concentration 𝑐𝑆:

𝑐𝑆 =�̇�𝑆

�̇�𝑆𝑢𝑠𝑝

(14)

�̇�𝑆 =�̇�𝑆

𝜌𝑆 (15)

𝜑𝑆 =�̇�𝑆

�̇�𝑆𝑢𝑠𝑝

=(�̇�𝑆𝜌𝑆

)

(�̇�𝑆𝑐𝑆

)=

𝑐𝑆𝜌𝑆

(16)

𝜑𝑆 =250

𝑘𝑔𝑚3

2600𝑘𝑔𝑚3

= 9.6 %

and the inflow pressure 𝑝𝑖:

𝑝𝑖 = 𝜌𝑆𝑢𝑠𝑝 ∙ 𝑔 ∙ ℎ (17)

𝜌𝑆𝑢𝑠𝑝 = ∑(𝜑𝑘

𝑘

∙ 𝜌𝑘) = 𝜑𝑆 ∙ 𝜌𝑆 + (1 − 𝜑𝑆) ∙ 𝜌𝐹 (18)

𝜌𝑆𝑢𝑠𝑝 = 1154 𝑘𝑔

𝑚3

𝑝𝑖 = 1154𝑘𝑔

𝑚3∙ 9.81

𝑚

𝑠2∙ 15 𝑚 = 170 𝑘𝑃𝑎

With the obtained results and the equations (5), (8) and (9) can be calculated the values in

table two.

table 2: solution table II

hydrocyclone

𝐷 in m

𝐷𝑜

𝐷𝑎

𝑉𝑜𝑉𝑎

𝜅

0.05 3.961 56.53 0.757

0.1 3.421 36.43 0.729

0.2 2.408 12.7 0.632

0.4 1.879 6.04 0.49

Lehrstuhl für Mechanische Verfahrenstechnik 5 Prof. Dr.-Ing. Jürgen Tomas

For the further hydrocyclone choice is important that the underflow suspension must be suffi-

ciently flow ability (solid maximum is about 𝜙𝑠,max = 35 Vol.-%). Thus, the volume ratio limits

the design of the hydrocyclone.

�̇�𝑜

�̇�𝑎≤ (

�̇�𝑜

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

(19)

(�̇�𝑜

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

= (�̇�𝑆𝑢𝑠𝑝 − �̇�𝑎

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

= (�̇�𝑆𝑢𝑠𝑝

�̇�𝑎− 1)

𝑙𝑖𝑚𝑖𝑡

(20)

By rewriting equation (14) results for �̇�𝑆𝑢𝑠𝑝:

�̇�𝑆𝑢𝑠𝑝 =�̇�𝑆

𝑐𝑆 (21)

The under flow �̇�𝑢 can be calculated with 𝜙𝑠,max.

�̇�𝑎 =�̇�𝐺

𝜙𝑠,max

=�̇�𝑆,𝐺

𝜌𝑄 ∙ 𝜙𝑠,max

(22)

Equation (21) and (22) in (20):

(�̇�𝑜

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

=�̇�𝑆 ∙ 𝜌𝑄 ∙ 𝜙𝑠,max

�̇�𝑆,𝐺 ∙ 𝑐𝑆− 1 (23)

The ratio of the mass flow rates of the underflow and the feed is called mass recovery of the

coarse. It shall be assumed, that all particles larger than the cut size (30 µm) are in the

coarse product.

𝜇𝐺 =�̇�𝑆,𝐺

�̇�𝑆= 1 − 𝑄3(𝑑𝑇) = 1 − 0.8 ∙ (

𝑑𝑇

𝑑80)𝑘

(24)

𝜇𝐺 = 1 − 0.8 ∙ (30

31.3)0.7

= 0.223

used in (23):

(�̇�𝑜

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

=𝜌𝑄 ∙ 𝜙𝑠,max

𝜇𝐺 ∙ 𝑐𝑆− 1 (25)

(�̇�𝑜

�̇�𝑎)

𝑙𝑖𝑚𝑖𝑡

=2650

𝑘𝑔𝑚3 ∙ 0.35

0.223 ∙ 250𝑘𝑔𝑚3

− 1 = 15.64

Lehrstuhl für Mechanische Verfahrenstechnik 6 Prof. Dr.-Ing. Jürgen Tomas

Only cyclones with a diameter of 𝐷 = 0.2 m and 𝐷 = 0.4 m are considered. Slim cyclones

have a higher separation sharpness, which can be seen in the calculations. Choosing of the

diameter 𝐷 = 0.2 m.

The suspension flow rate of a hydrocyclone can be calculated with equation (10) and (12)

with a cone angle of 𝛼 = 20°.

�̇�𝑍𝑦𝑘𝑙 =1

3.6∙ 0.04 𝑚 ∙ 0.06 𝑚 ∙ √

170000 𝑃𝑎

1154𝑘𝑔𝑚3

= 29.13𝑚3

ℎ

It follows for the required number of cyclones:

𝑛𝑍𝑦𝑘𝑙 =�̇�𝑆𝑢𝑠𝑝

�̇�𝑍𝑦𝑘𝑙

(26)

with using �̇�𝑆𝑢𝑠𝑝 which is calculated by equation (14):

�̇�𝑆𝑢𝑠𝑝 = 400𝑚3

ℎ (27)

𝑛𝑍𝑦𝑘𝑙 =400

𝑚3

ℎ

29.13𝑚3

ℎ

= 13.73 ⇒ 𝑛𝑔𝑒𝑤. = 14 (28)

14 cyclones with a diameter of 200 mm are required.